Question

A geo-stationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth. The time period of another satellite (in hours) at a height of 2R from the surface of the earth is [CBSE AIPMT 2012]

Moderate

Solution

From Kepler's third law, $T^{2} \propto r^{3}$
Hence, $T_{1}^{2} \propto r_{1}^{3} and T_{2}^{2} \propto r_{2}^{3}$
So, $\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{r_{2}^{3}}{r_{1}^{3}} = \frac{(3 R)^{3}}{(6 R)^{3}} or \frac{T_{2}^{2}}{T_{1}^{2}} = \frac{1}{8}$
$\Rightarrow T_{2}^{2} = \frac{1}{8} T_{1}^{2}$
$\Rightarrow T_{2} = \frac{24}{2 \sqrt{2}} = 6 \sqrt{2} h$

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